# 2.1. There are 30 possible choices for the first award,... etc., so the answer is 30 · 29 · 28 ·...

```2.1. There are 30 possible choices for the first award, then 29 for the second,
etc., so the answer is 30 &middot; 29 &middot; 28 &middot; 27 &middot; 26.
2.2. There is 107 possible phone numbers, and (in the same way as in the
previous problem) 10 &middot; 9 &middot; 8 &middot; 7 &middot; 6 &middot; 5 &middot; 4 phone numbers with all digits different.
Hence the probability is 10&middot;9&middot;8&middot;&middot;&middot;4
.
104
2.3. There are 6! = 720 ways to lineup. If we know that boys and girls
alternate and that the first is boy, then we know 3 places where boys are and 3
places for girls. Then the number of ways to put them is 3! times 3!. We get the
same number of ways if the first is girl. It follows that the number of ways to
put boys and girls so that they alternate is 2 &middot; 3! &middot; 3! = 72. Hence, the probability
is 72/720 = 1/10.
2.4. (a) The number of ways to choose the first committee is 50
8 . After
that the number of ways to choose the second committee is 42
7 . The number
42
50
of ways to choose two committees is 8 &middot; 7 .
(b) The number of ways to choose the first committee is the same as
before,
but the number of ways to choose the second committee is this time 50
7 . Hence
50
&middot;
.
the answer is 50
8
7
2 5 6
.
2.5. By the formula for the binomial distribution it is 82 16
1 60 35 24
24
=
2.6. The probability that you will have NO double 6 is 0 36
36
35 24
−λ λk
−24/36
.
The
Poisson
approximation
is
e
=
e
,
since
λ
=
np
=
24&middot;1/36
36
k!
and k = 0. The answer is 1 − e−24/36 .
2.7. In this case λ = 1, 500 &middot; 0.002, and k = 0, hence the probability is about
e−30 .
2.8. There are 10
4 = 210 ways to choose four people out of 5 couples. There
are 52 = 10 ways to choose two men and the same number of ways to choose
2
(5)
two women. Hence the probability is 210 = 100
210 .
(4)
2.9. Let Ai for i = 1, 2, 3, 4, 5 be the even that the electrician does not have
job on the weekday number i. We are asked to find the probability P (A1 ∪ A2 ∪
A3 ∪ A4 ∪ A5 ). It is equal to
X
i
P (Ai ) −
X
X
P (Ai ∩ Aj ) +
1≤i&lt;j≤4
X
P (Ai ∩ Aj ∩ Ak )−
1≤i&lt;j&lt;k≤4
P (Ai ∩ Aj ∩ Ak ∩ Al ) + P (A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5 ).
1≤i&lt;j&lt;k&lt;l≤4
The probability P (Ai ) is equal to 410 /510 , since there are 510 possible ways
for the customers to request days, and 410 ways to request the days so that the
day number i is free.
Similarly, P (Ai ∩ Aj ) = 310 /510 , P (Ai ∩ Aj ∩ Ak ) = 210 /510 , P (Ai ∩ Aj ∩
Ak ∩ Al ) = 1/510 and P (A1 ∩ A2 ∩ A3 ∩ A4 ∩ A5 ) = 0.
There are 5 summands in the first sum of the formula, 52 = 10 summands
in the second sum, 53 = 10 summands in the third sum, and 54 = 5 in the
fourth.
1
The answer is then
5&middot;
310
210
1
410
−
10
&middot;
+
10
&middot;
− 5 &middot; 10 .
510
510
510
5
2
```